Example: A rectangle
has its vertices on the $x$-axis, on the $y$-axis, at the
origin, and somewhere on the graph $y=4-x^2$ in the first
quadrant. Find the maximum possible area of such a
rectangle. Justify your answer.
Solution: As shown in the video, our rectangle has
width $x$ and height $y$, and so has area $xy$. But
$y=4-x^2$, so our area is $$A(x)= x(4-x^2) = 4x-x^3.$$ This
turns our word problem into just finding the maximum value
of $A(x)$ on $[0,2]$. Since $A'(x)=4-3x^2$, we have a
critical number when $4-3x^2=0$, or $x =
\frac{2}{\sqrt{3}}$. Then $y= 4-x^2 = 4 - \frac{4}{3} =
\frac{8}{3}$, and the area is $$A = xy =
\frac{16}{3\sqrt{3}} = \frac{16 \sqrt{3}}{9}.$$
(The video cuts out a few seconds too soon, but the end
of the calculation is shown to the left.)